answer:Solution: (1) The function f(x)= sqrt {1-x^{2}}, g(x)=a(x+b)(0 < aleqslant 1,bleqslant 0). We have y=f(x)cdot g(x)=a(x+b) sqrt {1-x^{2}}, (i) When b=0, f(x)cdot g(x)=ax sqrt {1-x^{2}}, -1leqslant xleqslant 1, Since f(-x)g(-x)=-ax sqrt {1-x^{2}}=-f(x)cdot g(x), the function y=f(x)g(x) is an odd function; (ii) When b < 0, f(x)cdot g(x)=a(x+b) sqrt {1-x^{2}}, -1leqslant xleqslant 1, Since f(- frac {1}{2})g(- frac {1}{2})=a(- frac {1}{2}+b)cdot frac { sqrt {3}}{2}, f( frac {1}{2})g( frac {1}{2})=a( frac {1}{2}+b)cdot frac { sqrt {3}}{2}, we have f(- frac {1}{2})g(- frac {1}{2})neq -f( frac {1}{2})g( frac {1}{2}) and f(- frac {1}{2})g(- frac {1}{2})neq f( frac {1}{2})g( frac {1}{2}), thus the function y=f(x)g(x) is neither odd nor even; (2) When b=0, the function y= frac {g(x)}{f^{2}(x)}= frac {ax}{1-x^{2}} is increasing on (-1,1). Reason: For any x_{1}, x_{2}, and -1 < x_{1} < x_{2} < 1, we have 1+x_{1}x_{2} > 0, (1-x_{1}^{2})(1-x_{2}^{2}) > 0, then y_{1}-y_{2}= frac {ax_{1}}{1-x_{1}^{2}}- frac {ax_{2}}{1-x_{2}^{2}}= frac {a(x_{1}-x_{2})(1+x_{1}x_{2})}{(1-x_{1}^{2})(1-x_{2}^{2})} < 0, we get y_{1} < y_{2}, thus the function y= frac {g(x)}{f^{2}(x)}= frac {ax}{1-x^{2}} is increasing on (-1,1). (3)h(x)=|af^{2}(x)- frac {g(x)}{a}|=|-ax^{2}-x+a-b|, the axis of symmetry is x=- frac {1}{2a}leqslant - frac {1}{2}, (i) When -1leqslant - frac {1}{2a}leqslant - frac {1}{2}, i.e., frac {1}{2}leqslant aleqslant 1, h(1)=|1+b|, h(-1)=|1-b|=1-b, h(- frac {1}{2a})=a+ frac {1}{4a}-b, h(x)_{max}=max{h(1),h(-1),h(- frac {1}{2a})}, a+ frac {1}{4a}-b is increasing when frac {1}{2}leqslant aleqslant 1, we have a+ frac {1}{4a}-bin[1-b, frac {5}{4}-b], thus we have h(x)_{max}=a+ frac {1}{4a}-b=2, we get a+b=2a+ frac {1}{4a}-2 is increasing when frac {1}{2}leqslant aleqslant 1, we have a+binleft[- frac {1}{2}, frac {1}{4}right]; (ii) When - frac {1}{2a} < -1, i.e., 0 < a < frac {1}{2}, h(x)_{max}=max{h(1),h(-1)}=1-b=2, thus b=-1, we get a+b=a-1in(-1,- frac {1}{2}). In conclusion, we have a+bin(-1,- frac {1}{4}]. Thus, the range of values for a+b is boxed{(-1,- frac {1}{4}]}.

answer:(1) Let the common ratio of the geometric sequence {a_n} be q. Given that a_1=frac{1}{2}, Since S_1+a_1, S_2+a_2, S_3+a_3 form an arithmetic sequence, we have: 2(S_{2}+a_{2})=S_{1}+a_{1}+S_{3}+a_{3} Rearranging, we get: S_{2}-S_{1}+2a_{2}=a_{1}+S_{3}-S_{2}+a_{3} This simplifies to: 3a_{2}=a_{1}+2a_{3} Substituting a_1 and a_2, we get: frac{3}{2}q = frac{1}{2} + q^2 Solving for q, we get q=1 or q=frac{1}{2}. Since {a_n} is a decreasing sequence, q=frac{1}{2}. Hence, a_{n}=a_{1}q^{n-1}=(frac{1}{2})^n (2) Given that b_n=a_nlog_2 2a_n=-ncdot(frac{1}{2})^n, We have T_{n}=-left[1cdot frac{1}{2}+2cdot (frac{1}{2})^{2}+ldots+(n-1)cdot (frac{1}{2})^{n-1}+ncdot (frac{1}{2})^{n}right] Multiplying both sides by frac{1}{2}, we get: frac{1}{2}T_{n}=-left[1cdot (frac{1}{2})^{2}+2cdot (frac{1}{2})^{3}+ldots+(n-1)cdot (frac{1}{2})^{n}+ncdot (frac{1}{2})^{n+1}right] Subtracting the two equations, we have: frac{1}{2}T_{n}=-left[frac{1}{2}+(frac{1}{2})^{2}+ldots+(frac{1}{2})^{n}-ncdot (frac{1}{2})^{n+1}right] This simplifies to: frac{1}{2}T_{n}=-frac{frac{1}{2}cdot left[1-(frac{1}{2})^{n}right]}{1-frac{1}{2}}+ncdot (frac{1}{2})^{n+1} Hence, T_{n}=(n+2)cdot (frac{1}{2})^{n}-2 And, frac{T_{n}+2}{n+2}=(frac{1}{2})^{n} From the inequality (frac{1}{2})^{n}geqslant frac{1}{16}, we get nleqslant 4. Hence, the maximum value of n is boxed{4}.

answer:First, calculate the probability of both balls being green: [ P(text{both green}) = frac{5}{10} cdot frac{10}{10+N} = frac{1}{2} cdot frac{10}{10+N} ] Then, calculate the probability of both balls being blue: [ P(text{both blue}) = frac{5}{10} cdot frac{N}{10+N} = frac{1}{2} cdot frac{N}{10+N} ] Adding these probabilities gives the total probability of drawing balls of the same color: [ frac{1}{2} cdot frac{10}{10+N} + frac{1}{2} cdot frac{N}{10+N} = 0.52 ] This equation simplifies to: [ frac{10+N}{10+N} = 1.04 ] Multiplying through by 10+N and simplifying: [ 10+N = 1.04(10+N) ] [ 10+N = 10.4 + 1.04N ] [ 10 - 10.4 = 1.04N - N ] [ -0.4 = 0.04N ] [ N = -0.4 / 0.04 = -10 ] As N cannot be negative, we need to recheck our calculation. The mistake is in setting up the equation for the equation simplification: [ frac{10 + N}{2(10+N)} = 0.52 ] [ 10 + N = 1.04(10+N) ] [ 10 + N = 10.4 + 1.04N ] [ 0.6 = 0.04N ] [ N = 0.6 / 0.04 ] [ N = 15 ] Thus, N = boxed{15}.

answer:The correct answer is boxed{text{D}}. Explanation is omitted.